A saturated solution of KHT in water will be prepared, meaning an aqueous soluti

Question

A saturated solution of KHT in water will be prepared, meaning an aqueous solution of K+and HT-with solid KHT also present. A known volume of the saturated solution will be titrated in the presence of phenolphthalein with a solution of sodium hydroxide of known concentration inorder to determine the concentration of HT-. Determination of the concentration of HT-is equivalent to determining the solubility of KHT in water, and we can then calculate Ksp. The solubility varies with temperature, so we are determining the ‘room temperature’ solubility.A literature value for the Kspof KHT is 1.16 x 10-3at 298 K as determined from solubility data (Journal of Chemical and Engineering Data2001, 46, 1362-1364, by P. Sousa and A. M. C. Lopes)

Procedure

1. Weigh out 1.5 grams of KHT and transfer it into a clean 400 mL beaker.

2. Add 150 mL of water to the beaker

3. Using your glass stirring rod, mix the contents by stirring for 15 minutes.

4. Using a dry piece of filter paper and a funnel, filter the solution into a second 400 mL

beaker.

5. Set up a 50 mL buret on a buret stand and rinse it through

with distilled water.

6. Fill the buret with the sodium hydroxide solution. Note the exact concentration of the

solution, which should be close to 0.04 M.

7. Record the initial buret volume reading on your data sheet.

8. Transfer exactly 30.0 mL of the saturated KHT solution into a clean 125 mL Erlenmeyer flask.

9. Add 3 drops of the phenolphthalein solution.

10. Titrate the saturated KHT solution with the sodium hydroxide solution until the pink color appears.

11. Record the final buret reading.

I'm lost on this I had posted the question before but I had no idea what was going on or where to put things

Name Data Titration Initial Buret Reading 50 m mL) soml ay.7 aag 23 y Final Buret Reading Calculations Moles OH added Moles HT consumed [HT] in original solution

Explanation / Answer

Reaction,

HT- + OH- ---> T^2- + H2O

moles HT- = moles OH-

Trial 1,

Volume NaOH added = 50 - 24.7 = 25.3 ml = 0.0253 L

moles of NaOH reacted = molarity x volume = 0.04 M x 0.0253 L = 0.001012 mol

moles of HT- present = 0.001012 mol

Volume of KHT solution taken = 30 ml = 0.030 L

molarity of [K+] = [HT-] in solution = moles/L = 0.001012 mol/0.030 L = 0.034 M

Volume of original KHT solution = 150 ml

molarity of [K+] = [HT-] in original solution = 0.034 M x 150 ml/30 ml = 0.17 M

Ksp = [K+][HT-] = (0.17)(0.17) = 2.89 x 10^-2

Similarly for other trial runs,

Trial 2,

Volume NaOH added = 50 - 22.9 = 27.1 ml = 0.0271 L

moles of NaOH reacted = molarity x volume = 0.04 M x 0.0271 L = 0.001084 mol

moles of HT- present = 0.001084 mol

Volume of KHT solution taken = 30 ml = 0.030 L

molarity of [K+] = [HT-] in solution = moles/L = 0.001084 mol/0.030 L = 0.036 M

Volume of original KHT solution = 150 ml

molarity of [K+] = [HT-] in original solution = 0.036 M x 150 ml/30 ml = 0.18 M

Ksp = [K+][HT-] = (0.18)(0.18) = 3.24 x 10^-2

Trial 3,

Volume NaOH added = 50 - 23.4 = 26.6 ml = 0.0266 L

moles of NaOH reacted = molarity x volume = 0.04 M x 0.0266 L = 0.001064 mol

moles of HT- present = 0.001064 mol

Volume of KHT solution taken = 30 ml = 0.030 L

molarity of [K+] = [HT-] in solution = moles/L = 0.001064 mol/0.030 L = 0.0355 M

Volume of original KHT solution = 150 ml

molarity of [K+] = [HT-] in original solution = 0.0355 M x 150 ml/30 ml = 0.18 M

Ksp = [K+][HT-] = (0.18)(0.18) = 3.24 x 10^-2

Average Ksp = 3.12 x 10^-2

%difference from literature value = (3.12 x 10^-2 - 1.16 x 10^-3) x 100/3.12 x 10^-2 = 96.3%

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